How many liters of a 25% saline solution must be added to 3 liters of a 10% saline solution to obtain a 15% saline solution?

Explanation

To solve the problem of how many liters of a 25% saline solution must be added to 3 liters of a 10% saline solution to obtain a 15% saline solution, we can set up an equation based on the concentration of saline in the solutions.

### Step-by-Step Solution:

1. **Define Variables:**
Let \( x \) be the number of liters of the 25% saline solution that needs to be added.

2. **Calculate the Amount of Saline in Each Solution:**
- The amount of saline in the 25% solution: \( 0.25x \) (since 25% of \( x \) liters is saline).
- The amount of saline in the 10% solution: \( 3 \times 0.1 = 0.3 \) liters (since 10% of 3 liters is saline).

3. **Total Volume of the Mixture:**
The total volume of the mixture after adding \( x \) liters of the 25% solution to 3 liters of the 10% solution is \( x + 3 \) liters.

4. **Set Up the Equation for the Final Concentration:**
We want the final solution to be 15% saline. Therefore, the total amount of saline in the final solution should equal 15% of the total volume:
\[ 0.25x + 0.3 = 0.15(x + 3) \]

5. **Expand and Rearrange the Equation:**
Expanding the right-hand side gives:
\[ 0.25x + 0.3 = 0.15x + 0.45 \]
Now, rearranging the equation:
\[ 0.25x - 0.15x = 0.45 - 0.3 \]
Simplifying this, we have:
\[ 0.10x = 0.15 \]

6. **Solve for \( x \):**
Dividing both sides by 0.10 gives:
\[ x = \frac{0.15}{0.10} = 1.5 \]

### Final Answer:
Thus, \( 1.5 \) liters of the 25% saline solution must be added.